   Heaven's Infinite Sume Here's the infinite sum for the heaven problem. You'll have to bear with the formatting here.... Sum from 1 to infinity of : (2^(n-1))/(3^n)  + (2^(n-1))/(3^n) * 2  The first few iterations look like this (add the days together): Day 1: 1/3 *1 + 1/3 *2 Day 2: 1/9 *1 + 1/9*1 + 1/9*2 + 1/9*2 == 2/9*1+2/9*2 Day 3: (1/27+1/27+1/27+1/27)*1 + (1/27+1/27+1/27+1/27)*2 == 4/27*1+4/27*2 Day N: 2^(n-1)/3^n*1+2^(n-1)/3^n*1 When you actually take the sum of the first 10 or so, you get about 3 days. It grows like this because for every extra day, there is a 1/3 chance (*2) from the day before you pick the wrong door again, but there are also exponentially more paths you can get to any given state, hence the 2^n-1. For example the first three days could look like: Day1 Day2 Day3 111 112 121 122 211 212 221 222 To pick a day 1 door on your third day, there is only 1/27 chance of it happeneing, but there are 4 different paths of three days which could lead you to this 4th day error. I hope this was moderately coherent. Just another kid Thursday, September 5, 2002 Recent Topics Fog Creek Home